Mastering Physics: Exercise 23.29

Mastering Physics: Exercise 23.29

A uniformly charged thin ring has radius 14.5 cm and total charge 22.5 nC . An electron is placed on the ring's axis a distance 33.0 cm from the center of the ring and is constrained to stay on the axis of the ring. The electron is then released from rest.

Part A

Question: Describe the subsequent motion of the electron.

Answer: When the electron is on either side of the center of the ring, the ring exerts an attractive force directed toward the center of the ring. This restoring force produces oscillatory motion of the electron along the axis of the ring, with amplitude, which equals initial distance between the electron and the ring's center. The force on the electron is not of the form F=-k*x so the oscillatory motion is not simple harmonic motion.

Part B

Question: Find the speed of the electron when it reaches the center of the ring.

Answer: 

U1 + K1 = U2 + K2

Initially K1 = 0

U1 − U2 = K2 = 1/2 × m_electron × (v_2)^2

U = qV = −eV

1/2 × m_electron × (v_2)^2 = −e (V1 − V2)

1/2 × m_electron × (v_2)^2 = e (V2 − V1)

v_2 = sqrt. (2e(V2 − V1) / (m_electron))

V1 = 1/(4πrϵ_0) × Q / sqrt.((x_1)^2 + R^2)

V1 = 8.99 ×10^(9) ×  22.5 ×10^(−9) / sqrt. ((0.330)^2 + (0.145)^2 )

V1 = 561.1718 V

V2 = 1/(4πrϵ_0) × Q / sqrt.((x_2)^2 + R^2)

V2 = 8.99 ×10^(9) ×  22.5 ×10^(−9) / sqrt. ((0.145)^2 )

V2 = 1395 V

v_2 = sqrt. (2e(V2 − V1) / (m_electron))

v_2 = sqrt. ( (2 × 1.602 ×10^(−9) × (1395 − 561.1778) / (9.12 ×10^(−31) )

v_2 =  1.71×10^7  m/s